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PS - Integers 2

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karthikpandian19: Legendary Member; Posts: 1665; Joined: Thu Nov 03, 2011 7:04 pmFollowed by: 70 members; Thanked: 165 times

by karthikpandian19 » Tue Dec 06, 2011 12:11 am

How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

(A) 31 (B) 51(C) 56(D) 62(E) 93

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Source: Beat The GMAT — Problem Solving | Tue Dec 06, 2011 12:11 am

neelgandham: Community Manager; Posts: 1060; Joined: Fri May 13, 2011 6:46 amFollowed by: 52 members; Location: Utrecht, The Netherlands; Thanked: 318 times

by neelgandham » Tue Dec 06, 2011 1:59 am

Hi Karthik,

The question has already been solved/discussed here https://www.beatthegmat.com/very-tricky- ... 25349.html

Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Dec 06, 2011 2:00 am

karthikpandian19 wrote:How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

(A) 31 (B) 51(C) 56(D) 62(E) 93

Sum of the digits is 5, so the numbers can be formed with the digits 0 to 5.

Numbers formed with digits 0, 0, 0 and 5 (0005) = 4!/3! = 4
Numbers formed with digits 0, 0, 1 and 4 (0041) = 4!/2! = 12
Numbers formed with digits 0, 0, 2 and 3 (0032) = 4!/2! = 12
Numbers formed with digits 0, 1, 1 and 3 (0113) = 4!/2! = 12
Numbers formed with digits 0, 1, 2 and 2 (0122) = 4!/2! = 12
Numbers formed with digits 1, 1, 1 and 2 (1112) = 4!/3! = 4

Therefore, number of positive integers less than 10,000 are there in which the sum of the digits equals 5 = (4 * 2) + (12 * 4) = 8 + 48 = 56

The correct answer is C.

Anurag Mairal, Ph.D., MBA
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